Question

Please answer it im having a test​

Answer 1

Step-by-step explanation:

$\frac{5 + \sqrt{2} }{5 - \sqrt{2} } = a + b \sqrt{2} \\ = > \frac{5 + \sqrt{2} }{5 - \sqrt{2} } \times \frac{5 + \sqrt{2} }{5 + \sqrt{2} } = a + b \sqrt{2} \\ = > \frac{ {(5 - \sqrt{2}) }^{2} }{ {(5)}^{2} - { \sqrt{(2)} }^{2} } = a + b \sqrt{2} \\ = > \frac{ {(5)}^{2} + 2 \times 5 \times \sqrt{2} + { \sqrt{(2)} }^{2} }{25 - 2} = a + b \sqrt{2} \\ = > \frac{25 + 10 \sqrt{2} + 2 }{23} = a + b \sqrt{2} \\ = > \frac{27 + 10 \sqrt{2} }{23}$

Comparing both sides, we get,

$a = \frac{27}{23}$

and,

$b \sqrt{2} = \frac{10 \sqrt{2} }{23} \\ = > b = \frac{10}{23}$

Answer 2

$\huge{ \underline{ \bold{ᴀɴsᴡᴇʀ....{ \heartsuit}}}}$

$$\begin{lgathered}\frac{5 + \sqrt{2} }{5 - \sqrt{2} } = a + b \sqrt{2} \\ = > \frac{5 + \sqrt{2} }{5 - \sqrt{2} } \times \frac{5 + \sqrt{2} }{5 + \sqrt{2} } = a + b \sqrt{2} \\ = > \frac{ {(5 - \sqrt{2}) }^{2} }{ {(5)}^{2} - { \sqrt{(2)} }^{2} } = a + b \sqrt{2} \\ = > \frac{ {(5)}^{2} + 2 \times 5 \times \sqrt{2} + { \sqrt{(2)} }^{2} }{25 - 2} = a + b \sqrt{2} \\ = > \frac{25 + 10 \sqrt{2} + 2 }{23} = a + b \sqrt{2} \\ = > \frac{27 + 10 \sqrt{2} }{23}\end{lgathered}$$

Comparing both sides, we get,

$$a = \frac{27}{23}$$

and,

$$\begin{lgathered}b \sqrt{2} = \frac{10 \sqrt{2} }{23} \\ = > b = \frac{10}{23}\end{lgathered}$$